2012-03-20 51 views
0

我需要一個PHP可以只顯示STORE_NAME在PHP打印只JSON字符串的一部分

我曾嘗試沒有成功

<?php 
require_once("admin/system/core.php"); 
$fetch = new server(); 
$fetch->connect(); 
$store = $fetch->getstore("movies"); 

print $store; 
?> 

的JSON看起來像這樣 [{"_id":{"$id":"4f67da1538fc5d7347000000"},"store_name":"movies","categories":{"name":"hoyts","products":{"name":"GoldClass","Price":"12.00","CashBack":"2.00"}}}]

+0

你試過了,你知道,解析它嗎? – 2012-03-20 06:08:14

+0

Post getStore()方法。 – Ben 2012-03-20 06:13:45

回答

2

以下我假設返回的值是json字符串。所以,你必須從字符串decode它的objectjson_decode功能

$store = $fetch->getstore("movies"); 
$data = json_decode($store); 

foreach($data as $d) { 
    echo $d->store_name; 
} 
1

假設你的陣列將

$array = array('id'=>'4f67da1538fc5d7347000000', 
     'store_name'=>'movies'); 
echo $json = json_encode($array); 

它呼應的

{"id":"4f67da1538fc5d7347000000","store_name":"movies"} 

那麼你可以使用

$data = json_decode($json); 
echo $data->store_name; 

which echo's movies